/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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NO

** BEGIN proof argument **
The following rule was generated while unfolding the analyzed TRS:
[iteration = 9] f(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f(s(s(s(s(s(s(s(s(id(_0))))))))),_1,_1)
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {} and theta2 = {_0->id(_0)}.
We have r|p = f(s(s(s(s(s(s(s(s(id(_0))))))))),_1,_1) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = f(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) loops w.r.t. the analyzed TRS.
** END proof argument **

** BEGIN proof description **
## Searching for a generalized rewrite rule
(a rule whose right-hand side contains a variable
that does not occur in the left-hand side)...
No generalized rewrite rule found!

## Applying the DP framework...
## 2 initial DP problems to solve.
## First, we try to decompose these problems into smaller problems.
## Round 1 [2 DP problems]:
  ## DP problem:
  Dependency pairs = [f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1)]
  TRS = {f(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1), id(s(_0)) -> s(id(_0)), id(0) -> 0}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... This DP problem is too complex! Aborting!
    ## Trying with lexicographic path orders... Failed!
    ## Trying with Knuth-Bendix orders... Failed!
  Don't know whether this DP problem is finite.
  ## DP problem:
  Dependency pairs = [id^#(s(_0)) -> id^#(_0)]
  TRS = {f(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1), id(s(_0)) -> s(id(_0)), id(0) -> 0}
    ## Trying with homeomorphic embeddings... Success!
  This DP problem is finite.
## A DP problem could not be proved finite.
## Now, we try to prove that this problem is infinite.
    ## Trying to find a loop (forward=true, backward=false, max=-1)
      # max_depth=10, unfold_variables=false:
        # Iteration 0: no loop found, 1 unfolded rule generated.
        # Iteration 1: no loop found, 1 unfolded rule generated.
        # Iteration 2: no loop found, 1 unfolded rule generated.
        # Iteration 3: no loop found, 1 unfolded rule generated.
        # Iteration 4: no loop found, 1 unfolded rule generated.
        # Iteration 5: no loop found, 1 unfolded rule generated.
        # Iteration 6: no loop found, 1 unfolded rule generated.
        # Iteration 7: no loop found, 1 unfolded rule generated.
        # Iteration 8: no loop found, 1 unfolded rule generated.
        # Iteration 9: success, found a loop, 1 unfolded rule generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^1.
        Let p1 = [0].
        We unfold the rule of L1 forwards at position p1
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L2 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(id(s(s(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^2.
        Let p2 = [0, 0].
        We unfold the rule of L2 forwards at position p2
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L3 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(id(s(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^3.
        Let p3 = [0, 0, 0].
        We unfold the rule of L3 forwards at position p3
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L4 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(id(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^4.
        Let p4 = [0, 0, 0, 0].
        We unfold the rule of L4 forwards at position p4
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L5 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(id(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^5.
        Let p5 = [0, 0, 0, 0, 0].
        We unfold the rule of L5 forwards at position p5
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L6 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(id(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^6.
        Let p6 = [0, 0, 0, 0, 0, 0].
        We unfold the rule of L6 forwards at position p6
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L7 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(s(id(s(s(_0))))))))),_1,_1) [unit] is in U_IR^7.
        Let p7 = [0, 0, 0, 0, 0, 0, 0].
        We unfold the rule of L7 forwards at position p7
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L8 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(s(s(id(s(_0))))))))),_1,_1) [unit] is in U_IR^8.
        Let p8 = [0, 0, 0, 0, 0, 0, 0, 0].
        We unfold the rule of L8 forwards at position p8
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L9 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(s(s(s(id(_0))))))))),_1,_1) [unit] is in U_IR^9.
  This DP problem is infinite.
Proof run on Linux version 3.10.0-1160.25.1.el7.x86_64 for amd64
using Java version 1.8.0_292
** END proof description **
Total number of generated unfolded rules = 19